louvig
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How to derive equation 3.6 in Covariant Physics by Emam
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It's a bit difficult to read. Also, perhaps needs some context re the author's notation.
louvig
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Sorry. I tried a screenshot from Kindle instead. I am able to click on it in my smartphone and make it full screen and is legible. The author is showing the covariance of classical mechanics using Einstein index notation. In this instance he is showing the transformation of the position vector which is straightforward and then the transformation of the derivative of the position vector. His point is to show ot transforms like a tensor and is therefore invariant.PeroK said:
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eq 3.6 has a typo, this index should read ##j## https://web.cortland.edu/moataz.emam/
The derivation is straight-forward:
Use that ##\hat{ \textbf{g}}_{i'} = \lambda^k_{i'} \hat{ \textbf{e}}_k ## and ##x^{i'} = \lambda^{i'}_j x^j##.
We get ## d\hat{ \textbf{g}}_{i'} = \hat{ \textbf{e}}_k d \lambda^k_{i'} ## and ##x^{i'} = x^j d\lambda^{i'}_j + \lambda^{i'}_j dx^j##.
And you will obtain the final step in that equation.
The derivation is straight-forward:
Use that ##\hat{ \textbf{g}}_{i'} = \lambda^k_{i'} \hat{ \textbf{e}}_k ## and ##x^{i'} = \lambda^{i'}_j x^j##.
We get ## d\hat{ \textbf{g}}_{i'} = \hat{ \textbf{e}}_k d \lambda^k_{i'} ## and ##x^{i'} = x^j d\lambda^{i'}_j + \lambda^{i'}_j dx^j##.
And you will obtain the final step in that equation.
louvig
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Thank you so much. Makes sense.
Moataz Emam
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louvig said:
Everything with primed coordinates was replaced with its transformation. So x’=lambda x and so on.